IE 423 Homework 1
Part 1
Part 1-1
Read image as a variable “img” and see the structure, dimension and class of it.
img <- readJPEG("C:/Users/Abdulsamed/Downloads/mutfakduvari.jpg")
imggray<-readJPEG("C:/Users/Abdulsamed/Downloads/grimutfakduvari.jpeg")
str(img)## num [1:512, 1:512, 1:3] 0.161 0.31 0.455 0.337 0.325 ...dim(img)## [1] 512 512 3class(img)## [1] "array"Part 1-2
This will display your image to test you read it correctly.
if(exists("rasterImage")){
plot(1:2, type='n', xlab = "", ylab = "")
rasterImage(img,1,1,2,2,interpolate = FALSE)
}
Make labels and margins smaller for multiple plot.
par(cex=0.5, mai=c(0.2,0.2,0.2,0.2))Define area for the images.
par(mar=rep(1,4))
par(fig=c(0,0.33,0.33,0.66))
image(img[,,1],col=hcl.colors(12, "Reds"),main="R Channel")
par(fig=c(0.33,0.66,0.33,0.66), new=TRUE)
image(img[,,2],col=hcl.colors(12, "Greens"),main="G Channel")
par(fig=c(0.66,0.99,0.33,0.66), new=TRUE)
image(img[,,3],col=hcl.colors(12, "Blues"),main="B Channel")
Grayscale
Make labels and margins smaller for multiple plot.
par(cex=0.5, mai=c(0.2,0.2,0.2,0.2))Define area for the images.
par(cex=0.5, mai=c(0.2,0.2,0.2,0.2))
par(fig=c(0,0.33,0.33,0.66))
image(img[,,1],col=gray.colors(20),main="R Channel")
par(fig=c(0.33,0.66,0.33,0.66), new=TRUE)
image(img[,,2],col=gray.colors(20),main="G Channel")
par(fig=c(0.66,0.99,0.33,0.66), new=TRUE)
image(img[,,3],col=gray.colors(20),main="B Channel")
Part 1-3
For each channel, take the average of the columns and plot the average as a line plot for each channel on a single plot.
par(cex=1, mai=c(0.2,0.2,0.2,0.2))
par(fig=c(0.1,0.9,0.1,0.9))
x1 <- 0
y1<- 0
x2 <- 0
y2<- 0
x3 <- 0
y3<- 0
for (i in 1:512){
x1[i] <-i
y1[i] <-mean(img[,i,1])
}
plot (x1,y1,type="l" , col ="red",main="Average of Columns for Each Channel", ylim=c(0.2,0.52), xlim=c(1,512))
for (i in 1:512){
x2[i]<-i
y2[i]<-mean(img[,i,2])
}
lines (x2,y2,type="l" , col ="green")
for (i in 1:512){
x3[i]<-i
y3[i]<-mean(img[,i,3])
}
lines (x3,y3,type="l" , col ="blue")
legend("topleft", c("R Channel","G Channel","B Channel"),fill=c("red","green","blue"))
Part 1-4
Subtract right half of the image from the left half for each channel.
imgnew <- array(0, dim = c(256,512,3))
imgnew[,,1]<-abs(img[1:256,,1]-img[257:512,,1])
imgnew[,,2]<-abs(img[1:256,,2]-img[257:512,,2])
imgnew[,,3]<-abs(img[1:256,,3]-img[257:512,,3])This will display your image to test if you read it correctly.
if(exists("rasterImage")){
plot(1:2, type='n', xlab = "", ylab = "")
rasterImage(imgnew,1,1,2,2,interpolate = FALSE)
}
Part 1-5
a
Apply median filtering to each channel of the image with window sizes 5x5
imgnew2 <-array(0,dim = c(512,512,3))
temporary <- array()
for(i in 3:510){
for(m in 3:510){
for(j in 1:3){
for(k in -2:2){
for(l in -2:2){
temporary <- c(temporary,img[i+k,m+l,j])
}
}
imgnew2[i,m,j] = median(temporary,na.rm=TRUE)
temporary<-array()
}
}
}
if(exists("rasterImage")){
plot(1:2, type='n', xlab = "", ylab = "")
rasterImage(imgnew2,1,1,2,2,interpolate = FALSE)
}
b
Apply median filtering to each channel of the image with window sizes 11x11
imgnew3 <-array(0,dim = c(512,512,3))
temporary <- array()
for(i in 6:507){
for(m in 6:507){
for(j in 1:3){
for(k in -5:5){
for(l in -5:5){
temporary <- c(temporary,img[i+k,m+l,j])
}
}
imgnew3[i,m,j] = median(temporary,na.rm=TRUE)
temporary<-array()
}
}
}
if(exists("rasterImage")){
plot(1:2, type='n', xlab = "", ylab = "")
rasterImage(imgnew3,1,1,2,2,interpolate = FALSE)
}
c
Apply median filtering to each channel of the image with window sizes 31x31 This code block lasts approximately 20 minute to execute since it includes too much loops.
imgnew4 <-array(0,dim = c(512,512,3))
temporary <- array()
for(i in 16:497){
for(m in 16:497){
for(j in 1:3){
for(k in -15:15){
for(l in -15:15){
temporary <- c(temporary,img[i+k,m+l,j])
}
}
imgnew4[i,m,j] = median(temporary,na.rm=TRUE)
temporary<-array()
}
}
}
if(exists("rasterImage")){
plot(1:2, type='n', xlab = "", ylab = "")
rasterImage(imgnew4,1,1,2,2,interpolate = FALSE)
}
As the window size increases the image begins to look blurry. Adjacent entries in the image matrix have common neighbor entries and if the window size increases, number of common neighbor entries increases. Since we replace each entry with the median of neighboring entries, adjacent entries become closer as the window size increases and it makes the image look blurry.
Part 2
Part 2-1
hist(imggray[,,1],breaks=20,main="Pixel Value Distribution",xlab="Pixel Values", ylab="Density",freq=FALSE)
It looks like a normal distribution even though it is slightly skewed to the right.
Part 2-2
fitdistr(imggray[,,1],"normal")## mean sd
## 0.3589119257 0.1784200553
## (0.0003484767) (0.0002464102)Our mean is: 0.3589119257, and our standard deviation is: 0.1784200553
Part 2-3
newz <-array(0,dim = c(512,512))
newz[,] <- imggray[,,1]
lowerlimit<-qnorm(0.0005, mean=0.3589119257, sd= 0.1784200553)
upperlimit<-qnorm(0.9995, mean=0.3589119257, sd= 0.1784200553)
for(x in 1:512){
for(y in 1:512){
if((newz[x,y]<lowerlimit)||(newz[x,y]>upperlimit))
{newz[x,y] = 0}
}
}
if(exists("rasterImage")){
plot(x=c(0,2.1),y=c(0,1), type='n', xlab = "", ylab = "")
rasterImage(newz,1.1,0,2.1,1,interpolate = FALSE)
rasterImage(imggray[,,1],0,0,1,1,interpolate=FALSE) }
It is quite difficult to spot a difference between the new and the previous image unless one zooms in. When zoomed in, artificial black pixels can be discerned as if they have been sprinkled from above. These dots give us an idea about the size of a single pixel and how its manipulation can alter the whole image. Changing the pixels that fall below the lower limit to black would not be discernible because they already are very close to black; however, the changes on the pixels that fall above the upper limit are quite visible because these pixels are usually surrounded by whitish dots. Namely, changing the pixels above the upper limit stands out more as defying the spontaneous pattern.
Part 2-4
Initial gray image
if(exists("rasterImage")){
plot(1:2, type='n', xlab = "", ylab = "", main = "Initial Gray Picture")
rasterImage(imggray,1,1,2,2,interpolate = FALSE)
}
i=array(1:512)
m=array(1:512)
j=array(1:3)
k=array(1:51)
l=array(1:51)
patch<- array(dim = c(51,51))
newimage<-array(imggray[,,1],dim = c(512,512))This part shows blackened image with 51 pixel steps from one patch to another.
for (i in seq(26,487,by=51)) {
for (m in seq(26,487,by=51)) {
for (k in -25:25) {
for (l in -25:25) {
patch[k+26,l+26]=imggray[i+k,m+l,1]
}
}
patchmean<-mean(patch[,])
patchsd<-sd(patch[,])
lowerlimit<-qnorm(0.0005, mean=patchmean, sd= patchsd)
upperlimit<-qnorm(0.9995, mean=patchmean, sd= patchsd)
for (n in -25:25) {
for (p in -25:25) {
if((patch[n+26,p+26]>upperlimit) | (patch[n+26,p+26]<lowerlimit)){
newimage[i+n,m+p]=0
}
}
}
}
}
if(exists("rasterImage")){
plot(1:2, type='n', xlab = "", ylab = "", main = "Blackened Image with 51 pixel steps")
rasterImage(newimage,1,1,2,2,interpolate = FALSE)
}
Plot the image obtained in Part3 and Part4 together
if(exists("rasterImage")){
plot(x=c(0,2.1),y=c(0,1), type='n', xlab = "", ylab = "")
rasterImage(newimage,1.1,0,2.1,1,interpolate = FALSE)
rasterImage(newz,0,0,1,1,interpolate = FALSE) }
In this part we assumed that the patches with particular size 51*51 distributed normally as we based our solutions in previous part. In each patch there exist some pixels which takes place beyond the limits of the pixel distribution but in this part we evaluate every patch seperately. So this method is robust to color change between seperate patches compared to the method we used in Part3. For that reason we expected the figure to have less black dots while taking patches for the main interest than taking the whole figure as our main interest. Observing figures obtained from Part3 and Part4 made our ideas strong. Moreover, if patches were taken larger, there would be more change in the image because at that way we took less amount of patches on the figure.